An aircraft flying at a horizontally height of 8000m above the ground at 200 metre per second drops a bomb on it's target on the ground. Find the angle between the vertical and the line joining the aircraft and the target immediately the bomb is released.

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To find the angle between the vertical and the line joining the aircraft and the target, we can use trigonometry.

Let's start by drawing a vertical line representing the aircraft's altitude of 8000m. Next, draw a horizontal line representing the ground. Now, draw a line connecting the aircraft with the target on the ground.

We know that the aircraft is flying at a horizontal speed of 200 meters per second, and we want to find the angle between the vertical (which is 90 degrees) and the line connecting the aircraft and the target.

Using trigonometry, we can calculate this angle as the inverse tangent of the opposite over adjacent sides of a right triangle. In this case, the opposite side is the vertical height of 8000m, and the adjacent side is the horizontal distance travelled by the aircraft.

Let's calculate the angle:

Tangent(angle) = Opposite / Adjacent
Tangent(angle) = 8000 / (200*t)

To find 't', we can calculate the time it takes for the bomb to reach the ground. Since the altitude is given as 8000m, and the vertical speed is the acceleration due to gravity (9.8 m/s^2), we can use the equation of motion:

s = ut + (1/2)at^2

Where:
s = distance
u = initial velocity
t = time
a = acceleration

Considering the initial velocity is zero because the bomb is dropped, the equation simplifies to:

s = (1/2)at^2

8000 = (1/2) * 9.8 * t^2

Simplifying further:

8000 = 4.9 * t^2
t^2 = 8000 / 4.9
t ≈ 25.89 seconds

Now, substitute this value of 't' back into the tangent equation:

Tangent(angle) = 8000 / (200 * 25.89)

Calculating this value using a calculator:

angle ≈ Tangent^-1 ( 8000 / (200 * 25.89) )

The result will give you the angle between the vertical and the line joining the aircraft and the target as approximately 1.16 degrees.